Answer:
B.
Step-by-step explanation:
Total surface area of the square pyramid
= 4 times the area of one triangle + Area of square
![= 4 \times \frac{1}{2} \times 6 \times 6 + {6}^{2} \\ \\ = 72 + 36 \\ \\ = 108 \: {in}^{2}](https://tex.z-dn.net/?f=%20%3D%204%20%5Ctimes%20%20%5Cfrac%7B1%7D%7B2%7D%20%20%5Ctimes%206%20%5Ctimes%206%20%2B%20%20%7B6%7D%5E%7B2%7D%20%20%5C%5C%20%20%5C%5C%20%20%3D%2072%20%2B%2036%20%5C%5C%20%20%5C%5C%20%20%3D%20108%20%5C%3A%20%20%7Bin%7D%5E%7B2%7D%20)
Answer:
9x+9
Step-by-step explanation
Combine like terms (6x and 3x, 2 and 7)
(3x2 + 9x + 6) − (8x2 + 3x − 10) + (2x + 4)(3x − 7)
First distribute the (2x+4)(3x-7) to get 6x2-2x-28
After this just add the like terms for all parts of the expression.
(3x2-8x2+6x2)+(9x-3x-2x)+(6+10-28)
x2+4x-12
So you answer is x2+4x-12
Given: ![f(x) = \frac{1}{x-2}](https://tex.z-dn.net/?f=f%28x%29%20%3D%20%5Cfrac%7B1%7D%7Bx-2%7D)
![g(x) = \frac{2x+1}{x}](https://tex.z-dn.net/?f=g%28x%29%20%3D%20%5Cfrac%7B2x%2B1%7D%7Bx%7D)
A.)Consider
![f(g(x))= f(\frac{2x+1}{x} )](https://tex.z-dn.net/?f=f%28g%28x%29%29%3D%20f%28%5Cfrac%7B2x%2B1%7D%7Bx%7D%20%29)
![f(\frac{2x+1}{x} )=\frac{1}{(\frac{2x+1}{x})-2}](https://tex.z-dn.net/?f=f%28%5Cfrac%7B2x%2B1%7D%7Bx%7D%20%29%3D%5Cfrac%7B1%7D%7B%28%5Cfrac%7B2x%2B1%7D%7Bx%7D%29-2%7D)
![f(\frac{2x+1}{x} )=\frac{1}{\frac{2x+1-2x}{x}}](https://tex.z-dn.net/?f=f%28%5Cfrac%7B2x%2B1%7D%7Bx%7D%20%29%3D%5Cfrac%7B1%7D%7B%5Cfrac%7B2x%2B1-2x%7D%7Bx%7D%7D)
![f(\frac{2x+1}{x} )=\frac{x}{1}](https://tex.z-dn.net/?f=f%28%5Cfrac%7B2x%2B1%7D%7Bx%7D%20%29%3D%5Cfrac%7Bx%7D%7B1%7D)
![f(\frac{2x+1}{x} )=1](https://tex.z-dn.net/?f=f%28%5Cfrac%7B2x%2B1%7D%7Bx%7D%20%29%3D1)
Also,
![g(f(x))= g(\frac{1}{x-2} )](https://tex.z-dn.net/?f=g%28f%28x%29%29%3D%20g%28%5Cfrac%7B1%7D%7Bx-2%7D%20%29)
![g(\frac{1}{x-2} )= \frac{2(\frac{1}{x-2}) +1 }{\frac{1}{x-2}}](https://tex.z-dn.net/?f=g%28%5Cfrac%7B1%7D%7Bx-2%7D%20%29%3D%20%5Cfrac%7B2%28%5Cfrac%7B1%7D%7Bx-2%7D%29%20%2B1%20%7D%7B%5Cfrac%7B1%7D%7Bx-2%7D%7D)
![g(\frac{1}{x-2} )= \frac{\frac{2+x-2}{x-2} }{\frac{1}{x-2}}](https://tex.z-dn.net/?f=g%28%5Cfrac%7B1%7D%7Bx-2%7D%20%29%3D%20%5Cfrac%7B%5Cfrac%7B2%2Bx-2%7D%7Bx-2%7D%20%7D%7B%5Cfrac%7B1%7D%7Bx-2%7D%7D)
![g(\frac{1}{x-2} )= \frac{x }{1}](https://tex.z-dn.net/?f=g%28%5Cfrac%7B1%7D%7Bx-2%7D%20%29%3D%20%5Cfrac%7Bx%20%7D%7B1%7D)
![g(\frac{1}{x-2} )= x](https://tex.z-dn.net/?f=g%28%5Cfrac%7B1%7D%7Bx-2%7D%20%29%3D%20x)
Since, ![f(g(x))=g(f(x))=x](https://tex.z-dn.net/?f=f%28g%28x%29%29%3Dg%28f%28x%29%29%3Dx)
Therefore, both functions are inverses of each other.
B.
For the Composition function ![f(g(x)) = f(\frac{2x+1}{x} )=x](https://tex.z-dn.net/?f=f%28g%28x%29%29%20%3D%20f%28%5Cfrac%7B2x%2B1%7D%7Bx%7D%20%29%3Dx)
Since, the function
is not defined for
.
Therefore, the domain is ![(-\infty,0)\cup(0,\infty)](https://tex.z-dn.net/?f=%28-%5Cinfty%2C0%29%5Ccup%280%2C%5Cinfty%29)
For the Composition function ![g(f(x)) =g(\frac{1}{x-2} )=x](https://tex.z-dn.net/?f=g%28f%28x%29%29%20%3Dg%28%5Cfrac%7B1%7D%7Bx-2%7D%20%29%3Dx)
Since, the function
is not defined for
.
Therefore, the domain is ![(-\infty,2)\cup(2,\infty)](https://tex.z-dn.net/?f=%28-%5Cinfty%2C2%29%5Ccup%282%2C%5Cinfty%29)